March 30, 2013

Below are a few images collected via Google Earth of significant elevation gains.

In North America two of the most significant slopes Mt McKinley in Alaska and Mt St Elias on the Alaska-Yukon border. For the former, the base-to-peak rise is something like 5500m while the latter is one of the closest big mountains to the ocean.

Mt McKinley, USA

Mt Saint Elias, Canada-USA

In Peru the Cotahuasi Canyon is one of the deepest canyons in the world. The following shows a rise of almost 5,500m up to Coropuna.

Cotahuasi Canyon to Coropuna, Peru

The remaining images are from Asia. For sheer rise it’s hard to go past Nanga Parbat in Pakistan. The following shows a gain of close to 7000 metres.

Nanga Parbat, Pakistan

The south-eastern slope of this mountain (Rupal face) has been called “the largest mountain face in the world” [1] rising some 4.5km in just 7km of horizontal distance!

Rupal face of Nanga Parbat, Pakistan

Finally we turn to the mighty Himalayas. There are numerous large elevation gains here – among the most notable are the area among the peaks of Dhaulagiri, Annapurna I and Machhapuchhare. The Kali Gandaki River flows through this region.

Like this:

March 19, 2013

This post is inspired by a question I asked myself during a recent one day international cricket game I watched on TV. It was pointed out that Pakistani batsman Nasir Jamshed had an impressive average of 50.26. I knew that he hadn’t played that many games so I was surprised that his average ended in .26 which suggests that he was dismissed more than just a few times. For example if his average were 50.22 I would have imagined that this was 50 and 2/9, so I could guess he had scored 452 runs and been dismissed 9 times.

So how many dismissals could he have had if his average were 50.26, assuming it is small? This was the way I thought about it mentally: the number 0.26 is just over 1/4, so it is probably of the form k/(4k-1), then it’s a question checking numbers of this form. To two decimal digits we have

2/7 = 0.29

3/11 = 0.27

4/15 = 0.27

5/19 = 0.26 (aha!)

Hence I guessed that he had been dismissed 19 times and had scored runs. Later on I checked that this was in fact the case. I thought it was cool that from a single average to two decimal places one could make a good guess of both the number of dismissals and the runs scored by the player (i.e. both the dividend and divisor), just based on the assumption that the ratio is simple. Of course if the decimal had been something like 50.25, it would have been more difficult to guess whether the he had scored 201, 402, 603, … runs for 4, 8, 12, … dismissals.

Below is a table showing for each value from 0.01 to 0.50 (in 0.01 increments) the simplest fraction that rounds to that value correct to two decimal places. Interestingly apart from fractions close to 0 and 0.5, the only case where the denominator exceeds 20 is for 0.34.

Decimal

Simplest ratio

Decimal

Simplest ratio

Decimal

Simplest ratio

Decimal

Simplest ratio

Decimal

Simplest ratio

0.01

1/67

0.11

1/9

0.21

3/14

0.31

4/13

0.41

7/17

0.02

1/41

0.12

2/17

0.22

2/9

0.32

6/19

0.42

5/12

0.03

1/29

0.13

1/8

0.23

3/13

0.33

1/3

0.43

3/7

0.04

1/23

0.14

1/7

0.24

4/17

0.34

10/29

0.44

4/9

0.05

1/19

0.15

2/13

0.25

1/4

0.35

6/17

0.45

5/11

0.06

1/16

0.16

3/19

0.26

5/19

0.36

4/11

0.46

6/13

0.07

1/14

0.17

1/6

0.27

3/11

0.37

7/19

0.47

7/15

0.08

1/12

0.18

2/11

0.28

5/18

0.38

3/8

0.48

10/21

0.09

1/11

0.19

3/16

0.29

2/7

0.39

7/18

0.49

17/35

0.10

1/10

0.20

1/5

0.30

3/10

0.40

2/5

0.50

1/2

To show how one of the above ratios is calculated (apart from trial and error), we will show as an example how to find fractions m/n whose decimal equivalent to two places is 0.26. In other words, or equivalently . We first find a fraction that is close to m/n, in this example 1/4 is appropriate. Using this, we set (more generally for the approximation p/q choose ) and find that or equivalently,

Hence for , ranges from 5 to 12, corresponding to the fractions (note that which rounds to 0.26 while which rounds down to 0.25).

For , ranges from 9 to 25, corresponding to the fractions . We can continue to find all fractions satisfying the above inequality but it is apparent that the simplest will correspond to and .